If the system of linear equations

$x+y+3 z=0$       $\ldots \ldots(\mathrm{i})$

$x+3 y+k^{2} z=0$    $\ldots . .(\mathrm{ii})$

$3 x+y+3 z=0$    ...........(iii)

$\left|\begin{array}{ccc}1 & 1 & 3 \\ 1 & 3 & k^{2} \\ 3 & 1 & 3\end{array}\right|=0$

$\Rightarrow 9+3+3 k^{2}-27-k^{2}-3=0$

$\Rightarrow \mathrm{k}^{2}=9$

(i) $-$ (iii) $\Rightarrow-2 \mathrm{x}=0 \Rightarrow \mathrm{x}=0$

Now from $(\mathrm{i}) \Rightarrow \mathrm{y}+3 \mathrm{z}=0$

$\Rightarrow \frac{y}{z}=-3$

$x+\frac{y}{z}=-3$